HCF MCQ Quiz - Objective Question with Answer for HCF - Download Free PDF
Last updated on Jun 7, 2025
Latest HCF MCQ Objective Questions
HCF Question 1:
Find the H.C.F. of p(x) = 2x3 – 3x2 – 2x + 3 and q(x) = 3x2 + 8x + 5.
Answer (Detailed Solution Below)
HCF Question 1 Detailed Solution
Given:
p(x) = 2x3 – 3x2 – 2x + 3 and q(x) = 3x2 + 8x + 5
Concept:
H.C.F. of two or more equations is the greatest factor that divides each of them exactly.
Calculation:
The factors of p(x) = 2x3 – 3x2 – 2x + 3
⇒ x2 × (2x – 3) – 1 × (2x – 3)
⇒ (x2 – 1) × (2x – 3)
⇒ (x – 1) × (x + 1) × (2x – 3)
And, the factors of q(x) = 3x2 + 8x + 5
⇒ 3x2 + 5x + 3x + 5
⇒ x × (3x + 5) + 1 × (3x + 5)
⇒ (3x + 5) × (x + 1)
∴ The required H.C.F. is (x + 1).HCF Question 2:
Find the HCF of \(4\frac{1}{5}, 5\frac{2}{7}\) and \(7\frac{4}{9}\)
Answer (Detailed Solution Below)
HCF Question 2 Detailed Solution
Given:
Numbers: \(4\frac{1}{5}\), \(5\frac{2}{7}\), \(7\frac{4}{9}\)
Formula Used:
HCF of fractions = HCF of numerators / LCM of denominators
Calculation:
Convert mixed fractions to improper fractions:
\(4\frac{1}{5}\) = (4 × 5 + 1)/5 = 21/5
\(5\frac{2}{7}\) = (5 × 7 + 2)/7 = 37/7
\(7\frac{4}{9}\) = (7 × 9 + 4)/9 = 67/9
Extract numerators and denominators:
Numerators: 21, 37, 67
Denominators: 5, 7, 9
Find HCF of numerators:
HCF(21, 37, 67) = 1 (since all are co-prime)
Find LCM of denominators:
LCM(5, 7, 9):
Prime factorization:
5 = 51
7 = 71
9 = 32
LCM = 51 × 71 × 32 = 315
HCF = HCF of numerators / LCM of denominators
HCF = 1 / 315
The HCF of \(4\frac{1}{5}\), \(5\frac{2}{7}\), and \(7\frac{4}{9}\) is 1/315 .
HCF Question 3:
A shopkeeper has 3 different qualities of Oil, 725 liters of 1st quality, 783 liters of 2nd quality and 812 liters of 3rd quality. Find the least possible number of bottles of equal size in which different Oil of different qualities can be filled without mixing?
Answer (Detailed Solution Below)
HCF Question 3 Detailed Solution
Given:
Oil quantities:
1st quality = 725 liters
2nd quality = 783 liters
3rd quality = 812 liters
Formula used:
Calculate the Highest Common Factor (HCF) of the given quantities to find the least possible number of bottles.
Calculation:
Find HCF of 725, 783, and 812.
Prime factorization:
725 = 5 × 5 × 29
783 = 3 × 3 × 3 × 29
812 = 2 × 2 × 7 × 29
Common factor = 29
Calculate the least possible number of bottles:
⇒ Total bottles = (725 ÷ 29) + (783 ÷ 29) + (812 ÷ 29)
⇒ Total bottles = 25 + 27 + 28 = 80
∴ The correct answer is option (4).
HCF Question 4:
A forester wants to plant 84 Neem trees, 126 Oak trees and 168 Banyan trees in equal rows (in terms of number of trees). Also, he wants to make distinct rows of trees (i.e. only one type of tree in one row). Find the number of rows (minimum) that are required.
Answer (Detailed Solution Below)
HCF Question 4 Detailed Solution
Given:
Number of Neem trees = 84
Number of Oak trees = 126
Number of Banyan trees = 168
Formula used:
To find the minimum number of rows required, calculate the HCF (Highest Common Factor) of the given numbers.
Minimum number of rows = HCF of 84, 126, and 168
Calculation:
Find the HCF of 84, 126, and 168
Prime factorization of 84 = 2 × 2 × 3 × 7
Prime factorization of 126 = 2 × 3 × 3 × 7
Prime factorization of 168 = 2 × 2 × 2 × 3 × 7
⇒ HCF = 2 × 3 × 7 = 42
Divide the total number of trees by the HCF to determine the number of rows for each type of tree.
Number of rows for Neem trees = 84 ÷ 42 = 2
Number of rows for Oak trees = 126 ÷ 42 = 3
Number of rows for Banyan trees = 168 ÷ 42 = 4
Add the rows for each type of tree.
Total rows = 2 + 3 + 4 = 9
∴ The correct answer is option (1).
HCF Question 5:
The greatest number which divides 68, 140 and 248 leaving the same remainder in each case is
Answer (Detailed Solution Below)
HCF Question 5 Detailed Solution
Given:
Numbers are 68, 140, and 248.
We need to find the greatest number that divides these numbers leaving the same remainder.
Formula Used:
When a number divides multiple numbers leaving the same remainder, the required number is the HCF (Highest Common Factor) of the differences of these numbers.
Calculation:
Find the differences:
140 - 68 = 72
248 - 140 = 108
248 - 68 = 180
Now, find the HCF of 72, 108, and 180.
Prime factorization:
72 = 23 × 32
108 = 22 × 33
180 = 22 × 32 × 5
Common factors: 22 × 32
HCF = 4 × 9 = 36
The greatest number that divides 68, 140, and 248 leaving the same remainder is 36.
Top HCF MCQ Objective Questions
Three piece of timber 143m, 78m and 117m long have to be divided into planks of the same length. What is the greatest possible length of each plank?
Answer (Detailed Solution Below)
HCF Question 6 Detailed Solution
Download Solution PDFGiven:
Length of timber1 = 143 m
Length of timber2 = 78 m
Length of timber3 = 117 m
Calculation:
Greatest possible length of each plank = HCF of 143, 78 and 117
143 = 13 × 11
78 = 13 × 2 × 3
117 = 13 × 3 × 3
HCF is 13
∴ Greatest possible length of each plank is 13 m.
The sum of two numbers is 288 and their HCF is 16. How many pairs of such numbers can be formed?
Answer (Detailed Solution Below)
HCF Question 7 Detailed Solution
Download Solution PDFGiven:
The sum of two numbers is 288 and their HCF is 16
Calculations:
Let the ratio of number be x : y
So the numbers will be 16x & 16y (HCF is an integral part of a number)
According to the question
16x + 16y = 288
⇒ 16(x + y) = 288
⇒ x + y = 18
Pairs of x, y can be (1, 17) (5, 13) (7, 11)
So there can be only 3 pairs.
∴ The correct choice is option 1.
Two numbers are in the ratio 7 : 11. If their HCF is 28, then sum of the two numbers is:
Answer (Detailed Solution Below)
HCF Question 8 Detailed Solution
Download Solution PDFGiven:
Ratio of numbers = 7 ∶ 11
HCF = 28
Calculation:
Let the numbers be 7x and 11x
HCF of 7x and 11x is x
HCF = x = 28
The numbers will be 7 × 28 and 11 × 28
⇒ The numbers will be 196 and 308
Sum of numbers = 196 + 308
⇒ Sum of numbers = 504
∴ Sum of numbers is 504
Shortcut Trick Note that the sum of two numbers is asked.
Let the numbers be 7x and 11x.
Add the numbers:
⇒ 7x + 11x
⇒ 18x
Now see, the final number must be the multiple of 18, so in options only 504 is multiple of 18.
∴ The sum of two number is 504.
Find the HCF of (4315 − 1) and (425 − 1).
Answer (Detailed Solution Below)
HCF Question 9 Detailed Solution
Download Solution PDFGiven:
(4315 − 1) and (425 − 1)
Concept used:
HCF of (am − 1) and (an − 1) is (aHCF(m,n) − 1).
Calculations:
HCF(315, 25) = 5
According to the concept,
HCF {(4315 − 1), (425 − 1)}
= (4HCF(315, 25) − 1)
= (45 − 1)
= 1024 − 1
= 1023
Hence, The Required value is 1023.
The H.C.F. of (x3 + x2 + x + 1) and (x4 – 1) is
Answer (Detailed Solution Below)
HCF Question 10 Detailed Solution
Download Solution PDFGiven:
The H.C.F. of (x3 + x2 + x + 1) and (x4 – 1) is
Calculation:
⇒ (x3 + x2 + x + 1) = x2(x + 1) + 1(x + 1)
⇒ (x + 1) (x2 + 1)
⇒ x4 – 1 = (x2 – 1) (x2 + 1)
⇒ (x + 1) (x – 1) (x2 + 1)
∴ Required HCF is (x + 1) (x2 + 1)
The product of two numbers is 1521 and the HCF of these numbers is 13. Find the number of such pairs?
Answer (Detailed Solution Below)
HCF Question 11 Detailed Solution
Download Solution PDFGIVEN:
The product of two numbers is 1521 and the HCF of these numbers is 13.
CONCEPT:
HCF: The highest common factor (HCF) is found by finding all common factors of two numbers and selecting the largest one.
CALCULATION:
Suppose the numbers are 13a and 13b as the HCF of these numbers is 13.
We can write:
13a × 13b = 1521
⇒ ab = 9
∴ Only possible pair is 13, 117
Mistake PointsAccording to question,
ab = 9
For a = 1 and b = 9
The numbers will be 13 and 117 and their HCF will be 13
Here we will not consider a = 3 and b = 3.
The numbers will be 39 and 39.
Here HCF would be 39 which does not satisfy the given condition.
The sum of two positive numbers is 240 and their HCF is 15. Find the number of pairs of numbers satisfying the given condition.
Answer (Detailed Solution Below)
HCF Question 12 Detailed Solution
Download Solution PDFGiven:
The sum of two number positive number is 240 and their HCF is 15.
Calculation:
Let two positive number is 15x and 15y where x and y should be coprime that means x and y should have HCH as 1.
According to the question
The sum of the number is
⇒ 15x + 15y = 240
⇒ x + y = 16
Now, we have to find the number of pair in which sum of the two number is 16 but no common factor between them, such pair is
⇒ (1, 15) (3, 13) (5, 11) (7, 9)
∴ Total possible pairs is 4.
Confusion Points
We can't take (2, 14), (4, 12), (6, 10), (8, 8) Because In these cases the pair should be co-prime.
The product of the two numbers is 2160 and their HCF is 12. The number of possible pairs of such numbers is:
Answer (Detailed Solution Below)
HCF Question 13 Detailed Solution
Download Solution PDFGiven:
The product of the two numbers is 2160 and their HCF is 12.
Calculation:
Let the number be x and y.
According to the question,
12x × 12y = 2160
⇒ 144xy = 2160
⇒ xy = \(\dfrac{2160}{144}\) = 15
⇒ xy = 15
Possible pair = (1 × 15) and (3 × 5)
∴ The number of possible pairs of such numbers is 2.
HCF of two number is 4 and the sum of those two numbers is 36. Find how many such pair of number is possible.
Answer (Detailed Solution Below)
HCF Question 14 Detailed Solution
Download Solution PDFGiven:
HCF of two number is 4 and the sum of those two numbers is 36.
Concept Used:
Concept of HCF
HCF is the least common factor among two or more numbers.
Calculation:
HCF of two number is 4
Let, those numbers be 4x and 4y where x and y are prime to each other
Accordingly,
4x + 4y = 36
⇒ 4(x + y) = 36
⇒ (x + y) = 9
Now,
9 = 8 + 1
9 = 7 + 2
9 = 6 + 3
9 = 5 + 4
In all these cases only (8,1); (7,2); and (5,4) are prime to each other. So, such three pair is possible
∴ Such three pair of number is possible.
Three number are in the ratio of 4 ∶ 5 ∶ 6, and their LCM is 180. Their HCF is:
Answer (Detailed Solution Below)
HCF Question 15 Detailed Solution
Download Solution PDFGiven
Ratios of numbers: 4 : 5 : 6
LCM of the numbers: 180
Concept:
When three numbers are in the ratio a:b:c, and their LCM is n, then their HCF is n/(abc).
Solution:
HCF of the numbers = 180/LCM of (4, 5 and 6) = 180/60 ⇒ 3
Therefore, their HCF is 3.
Alternate Method
Ratio of numbers = 4 : 5 : 6
Let numbers be 4x, 5x and 6x where x is the HCF of numbers.
So
LCM of (4x, 5x, 6x) = 60x
60x = 180
so
x = 3 = HCF of numbers
So 3 is the correct answer.